Use the information given about the angle \( \theta, \cot \theta=-4, \sec \theta<0,0 \leq \theta<2 \pi \), to find the exact values of the following. (a) \( \sin (2 \theta) \), (b) \( \cos (2 \theta) \), (c) \( \sin \frac{\theta}{2} \), and (d) \( \cos \frac{\theta}{2} \) (a) \( \sin (2 \theta)=-\frac{8}{17} \) (Type an exact answer, using radicals as needed.) (b) \( \cos (2 \theta)=\frac{15}{17} \) (Type an exact answer, using radicals as needed.) (c) \( \sin \frac{\theta}{2}=\square \) (Type an exact answer, using radicals as needed.)

To solve the problem, we start with the given information about the angle \( \theta \): 1. \( \cot \theta = -4 \) 2. \( \sec \theta < 0 \) 3. \( 0 \leq \theta < 2\pi \) ### Step 1: Determine the quadrant of \( \theta \) Since \( \cot \theta = -4 \), this means that \( \tan \theta = -\frac{1}{4} \). The cotangent is negative in the second and fourth quadrants. However, since \( \sec \theta < 0 \), which implies that \( \cos \theta < 0 \), we conclude that \( \theta \) must be in the second quadrant. ### Step 2: Find \( \sin \theta \) and \( \cos \theta \) Using the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can express \( \cos \theta \) and \( \sin \theta \) in terms of a right triangle. Let: - \( \cos \theta = -4k \) - \( \sin \theta = k \) From the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ k^2 + (-4k)^2 = 1 \] \[ k^2 + 16k^2 = 1 \] \[ 17k^2 = 1 \] \[ k^2 = \frac{1}{17} \] \[ k = \frac{1}{\sqrt{17}} \quad (\text{since } k > 0 \text{ in the second quadrant}) \] Thus, we have: \[ \sin \theta = k = \frac{1}{\sqrt{17}}, \quad \cos \theta = -4k = -\frac{4}{\sqrt{17}} \] ### Step 3: Calculate \( \sin(2\theta) \) and \( \cos(2\theta) \) Using the double angle formulas: \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] Calculating \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \left(\frac{1}{\sqrt{17}}\right) \left(-\frac{4}{\sqrt{17}}\right) = -\frac{8}{17} \] Calculating \( \cos(2\theta) \): \[ \cos(2\theta) = \left(-\frac{4}{\sqrt{17}}\right)^2 - \left(\frac{1}{\sqrt{17}}\right)^2 = \frac{16}{17} - \frac{1}{17} = \frac{15}{17} \] ### Step 4: Calculate \( \sin\left(\frac{\theta}{2}\right) \) and \( \cos\left(\frac{\theta}{2}\right) \) Using the half-angle formulas: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}} \] \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} \] Calculating \( \sin\left(\frac{\theta}{2}\right) \): \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{4}{\sqrt{17}}\right)}{2}} = \sqrt{\frac{1 + \frac{4}{\sqrt{17}}}{2}} = \sqrt{\frac{\frac{\sqrt{17} + 4}{\sqrt{17}}}{2}} = \sqrt{\frac{\sqrt{17} + 4}{2\sqrt{17}}} \] Calculating \( \cos\left(\frac{\theta}{2}\right) \): \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \left(-\frac{4}{\sqrt{17}}\right)}{2}} = \sqrt{\frac{1 - \frac{4}{\sqrt{17}}}{2}} = \sqrt{\frac{\frac{\sqrt{17} - 4}{\sqrt{17}}}{2}} = \sqrt{\frac{\sqrt{17} - 4}{2\sqrt{17}}} \] ### Final Answers (a) \( \sin(2\theta) = -\frac{8}{17} \) (b) \( \cos(2\theta) = \frac{15}{17} \) (c) \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{17} + 4}{2\sqrt{17}}} \) (d) \( \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{17} - 4}{2\sqrt{17}}} \)